Product of inertia and principal axes

For a plane area, the product of inertia about a chosen $x$–$y$ pair is

where: $x,y$ are the coordinates of an area element $dA$ measured from the chosen axes. Unlike $I_x=\int y^2\,dA$ and $I_y=\int x^2\,dA$, the integrand $xy$ can be positive in one quadrant and negative in another, so $I_{xy}$ itself can come out positive, negative, or exactly zero — its sign is physically meaningful and must never be "corrected" away.

Symmetry kills it instantly. If one of the reference axes coincides with an axis of symmetry of the area, $I_{xy}=0$ about that pair, because every element at $(x,y)$ is matched by a mirror element at $(x,-y)$ whose $xy$ contribution exactly cancels. The trap is assuming this for any symmetric shape — the symmetry axis must actually be one of the chosen reference axes; the same symmetric shape referred to arbitrarily tilted axes generally has $I_{xy}\ne0$.

Transferring between axis sets works like the familiar parallel-axis theorem, but with a mixed term:

where: $\bar I_{xy}$ is the product of inertia about centroidal axes parallel to $x,y$, $A$ is the area, and $\bar x,\bar y$ are the centroid's coordinates relative to the reference axes (carrying their own sign). For a composite area, sum the transferred $I_{xy}$ of each piece, sign included.

Rotating axes by an angle $\theta$ (measured from the centroidal $x$-axis, positive counterclockwise) mixes $I_x$, $I_y$, and $I_{xy}$ into new values $I_{x'}$, $I_{y'}$, $I_{x'y'}$:

where: $I_x,I_y,I_{xy}$ are known values about a centroidal axis pair, and the primed quantities are the corresponding values about axes rotated by $\theta$. These three equations have the same algebraic skeleton as plane-stress transformation, which is why the same graphical tool — Mohr's circle — applies here: plotting the points $(I_x,I_{xy})$ and $(I_y,-I_{xy})$ and drawing the circle through them gives $I_{max}$, $I_{min}$, and $\theta_p$ without touching the transformation equations directly.

Think of $I_{x'y'}(\theta)$ as a continuous function of the rotation angle: it starts at whatever $I_{xy}$ the reference axes give, oscillates through the transformation equation as $\theta$ sweeps around, and by the intermediate-value property must cross zero somewhere in every $90^\circ$ window (since $I_{x'}$ and $I_{y'}$ are simultaneously extremal exactly there). Setting $I_{x'y'}=0$ in the transformation equation pins that crossing:

and substituting $\theta_p$ back into $I_{x'}$, $I_{y'}$ collapses the square-root combination that gives the principal moments of inertia:

where: $I_{max}$ and $I_{min}$ are the extreme moments of inertia at the principal orientation, and the two roots of $\tan2\theta_p$ (90 degrees apart) locate the two perpendicular principal axes — one carries $I_{max}$, the other $I_{min}$. A useful check on any principal-axis calculation is that $I_{max}+I_{min}=I_x+I_y$ always, because that sum is the polar moment $J$, which does not change under rotation.

A cross-section has, about its centroidal $x$–$y$ axes, $I_x = 8\times10^6\text{ mm}^4$, $I_y = 3\times10^6\text{ mm}^4$, $I_{xy} = 2\times10^6\text{ mm}^4$. Find the principal moments of inertia and the orientation of the principal axes.

Step 1. Apply the angle formula: $\tan2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{-2(2\times10^6)}{8\times10^6-3\times10^6} = \dfrac{-4\times10^6}{5\times10^6} = -0.8$.

Step 2. Solve for the angle: $2\theta_p = \tan^{-1}(-0.8) = -38.66^\circ$, so $\theta_p = -19.33^\circ$. The second principal axis sits $90^\circ$ away, at $\theta_p = -19.33^\circ+90^\circ = 70.67^\circ$.

Step 3. Compute the average and the radius term: $I_{avg} = \frac{I_x+I_y}{2} = \frac{8+3}{2}\times10^6 = 5.5\times10^6$, and $R = \sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} = \sqrt{(2.5\times10^6)^2+(2\times10^6)^2} = \sqrt{6.25\times10^{12}+4\times10^{12}} = \sqrt{1.025\times10^{13}} = 3.20\times10^6$.

Step 4. Combine: $I_{max} = 5.5\times10^6+3.20\times10^6 = 8.70\times10^6\text{ mm}^4$ and $I_{min} = 5.5\times10^6-3.20\times10^6 = 2.30\times10^6\text{ mm}^4$. Check the invariant: $I_{max}+I_{min}=11.0\times10^6 = I_x+I_y$ ✓.

The principal moments are $I_{max}=8.70\times10^6\text{ mm}^4$ and $I_{min}=2.30\times10^6\text{ mm}^4$, oriented at $\theta_p=-19.33^\circ$ and $70.67^\circ$.

Competitive stretch

Where would $I_{xy}=2\times10^6\text{ mm}^4$ itself come from for a real shape? A thin right triangle with legs $b=60$ mm, $h=90$ mm along the axes has $I_{xy,O}=\dfrac{b^2h^2}{24}=\dfrac{60^2\times90^2}{24}=1.215\times10^6\text{ mm}^4$ about the right-angle corner. Its centroid sits at $\bar x=b/3=20$ mm, $\bar y=h/3=30$ mm, with $A=\tfrac12bh=2700\text{ mm}^2$, so the parallel-axis relation gives the centroidal value $\bar I_{xy}=I_{xy,O}-A\bar x\bar y = 1.215\times10^6-2700(20)(30) = 1.215\times10^6-1.62\times10^6 = -4.05\times10^5\text{ mm}^4$ — negative, and that sign must be carried unchanged into any later rotation calculation for this triangle.

• Treating a negative $I_{xy}$ as a mistake to fix — the sign is physically real and must be carried through every parallel-axis and transformation step.
• Assuming $I_{xy}=0$ for any symmetric shape — the symmetry axis must coincide with one of the chosen reference axes; tilt the axes and the same shape generally has $I_{xy}\ne0$.
• Sign-dropping in $A\bar x\bar y$ — the centroid coordinates $\bar x,\bar y$ keep their signs relative to the reference axes, including negative values when the centroid lies in a "negative" quadrant.
• Mixing this topic's scope (finding principal axes of an area) with unsymmetric-bending stress analysis — that application belongs to a later flexure chapter and is not tested here.

Ask the tutor: if the direction of $\theta_p$ or the sign convention in the rotation equations is confusing, ask the AI tutor to walk through this worked example with a labeled sketch.